Tic tac toe confusion

[Naughty Albatross]

> Don't we need a return false statement?
The first thing that you need to do is to check for all winning possibilities. After that, you may place a "return false" at the end of declareWinner() function. "return true" means a winner has been declared and vice-versa.

[Naughty Albatross]

if(ticTac[0][0] == "X")
    cout_this << "X won!";
    if(ticTac[0][0] == "O")
    cout_this << "O won!";

Can you rewrite this piece of code in just a single cout statement?

[FBHSIE]

You say all possibilities? Wouldn't that mean I'm going to be writing quite a bit of if statements or am I misinterpreting?

For each subscript combination, wouldn't I have to write an if-then statement as displayed previously?

(sorry your comment didn't load)

I never have before, so I'm going to say no.

I've always done it as you've shown for every program I've ever made.

[Naughty Albatross]

So can you answer the question above?

You are actually learning how to check for a winning possibility the proper way.

[FBHSIE]

The one about rewriting the code into a single cout statement?

Well,

if(ticTac[0][0] == "X")
    cout_this << "X won!";
    if(ticTac[0][0] == "O")
    cout_this << "O won!";

Makes sense to me as this is how I've always done it, but I can attempt to.

Why are we trying to, though? Is there something wrong with what I'm trying to do?

The cout statements just display who wins

[Naughty Albatross]

> Why are we trying to, though?
It is a must, though the code is not wrong.

[FBHSIE]

And the only thing I have to figure out is how to put the two cout_this together on one line without changing what's inside ()?

[Naughty Albatross]

The goal : There are four statements in total in this piece of code. Remove all if-statements (two if-statements) and one cout statement, leaving only one cout statement.

[FBHSIE]

(ticTac[0][0] == "X")

(ticTac[0][0] == "O")
cout_this << "O won!";

Well this is awfully an interesting experiment so far.

[Naughty Albatross]

You give yourself another ten minutes. Let me know if you finally give up.

Note : The new piece of code must retain the same meaning the old piece of code has.

[FBHSIE]

I have no idea why this is so hard to do.

One cout statement same function

[Naughty Albatross]

if (ticTac[0][0]==ticTac[0][1] && ticTac[0][0] ==ticTac[0][2] && ticTac[0][0] != "*")
{
     if(ticTac[0][0] == "X")
    cout_this << "X won!";
    else
    cout_this << "O won!"; // (1)

     return true;
}

You must have to declare the player X or the player O as the winner soon or later. There is one way that allows you to do it in a single cout statement :

if (ticTac[0][0]==ticTac[0][1] && ticTac[0][0] ==ticTac[0][2] && ticTac[0][0] != "*")
{
     cout << "Congratulations player " << ticTac[0][0] << ", you won!." << "\n\n"; // (2)

     return true;
}

[FBHSIE]

Later? Well, alright.

What does this display exactly? ticTac[0][0]

How would that show what player won?

(when I say player I meant you put it congratulations player * * as if it will show)

(I can't currently test it on my computer thus why I'm asking)

[Naughty Albatross]

A tic tac toe board's square may contain one of the three values : asterisk ("*"), X ("X") or ("O").

For example :
ticTac[1][0] = "O";
cout << "I am " << ticTac[1][0];

Output :

I am O 

In the case you mentioned, cout << ticTac[0][0] can output either "X" or "O". And don't worry, all the three squares are equal.

And there is no asterisk ("*"), because ticTac[0][0] has been very strictly guaranteed along with ticTac[0][1], and ticTac[0][2] that it can't be an asterisk.

[FBHSIE]

Thanks for the example. I understand now.

[Naughty Albatross]

if (ticTac[0][0]==ticTac[0][1] && ticTac[0][0] ==ticTac[0][2] && ticTac[0][0] != "*")
{
    cout << "Congratulations player " << ticTac[0][0] << ", you won!." << "\n\n";

     return true;
}

Congrats. You have successfully checked for a winning possibility.

[Naughty Albatross]

> You say all possibilities?
Look at this, you will understand.

1.
X X X
* * *
* * *

2.
* * *
X X X
* * *

3.
* * *
* * *
X X X

4.
X * *
X * *
X * *

5.
* X *
* X *
* X *

6.
* * X
* * X
* * X

7.
X * *
* X *
* * X

8.
* * X
* X *
X * *

[Naughty Albatross]

So can you tell which one is better?

if (ticTac[0][0]=="X" && ticTac[0][1] == "X" && ticTac[0][2] == "X")
{
     cout << "Congratulations player X, you won!." << "\n\n";

     return true;
}

if (ticTac[0][0]=="O" && ticTac[0][1] == "O" && ticTac[0][2] == "O")
{
     cout << "Congratulations player O, you won!." << "\n\n";

     return true;
}

And :

if (ticTac[0][0]==ticTac[0][1] && ticTac[0][0] ==ticTac[0][2] && ticTac[0][0] != "*")
{
     cout << "Congratulations player " << ticTac[0][0] << ", you won!." << "\n\n";

     return true;
}

[FBHSIE]

I apologize for falling asleep.

That makes sense, but something tells me you don't want me to write eight if-else statements.

(note I decided I'm going to write the eight if-then statement)

[AbstractionAnon]

something tells me you don't want me to write eight if-else statements.

That's certainly one way to do it.

Consider this:

struct Combo
{   int row;
    int col;
};
    
bool check_winner (string tictac[3][3], string who)
{  const Combo combo[8][3] = 
   {  0,0, 0,1, 0,2,    //  Row 0
      1,0, 1,1, 1,2,    //  Row 1
      2,0, 2,1, 2,2,    //  Row 2
      0,0, 1,0, 2,0,    //  Col 0
      1,0, 1,1, 2,1,    //  Col 1
      2,0, 2,1, 2,2,    //  Col 2
      0,0, 1,1, 2,2,    //  LR Diagonal
      0,2, 1,1, 2,0,    //  RL Diagonal
    };
    
    for (int i=0; i<8; i++) //  Check each combo
    {   int count = 0;
        for (int j=0; j<3; j++)     // Check row and col
        {   if (tictac[combo[i][j].row][combo[i][j].col] == who)
                count++;  //  One cell matched
        }
        if (count == 3)
            return true;    //  All three cells matched
    }
    return false;   //  No combo's matched
}